3 Marks Question - Physics STD 11 Science Questions - Vidyadip

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Question 13 Marks

The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 107K, and its outer surface at a temperature of about 6000K. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases? Check if your guess is correct from the following data : mass of the Sun = 2.0 × 1030kg, radius of the Sun = 7.0 × 108m.

Answer

Mass of the Sun, M = 2.0 × 10³⁰kg

Radius of the Sun, R = 7.0 × 10⁸m

Density, $\rho= ?$

$\rho=\frac{\text{mass}}{\text{volume}}=\frac{\text{M}}{\frac{4}{3}\pi\text{R}^3}=\frac{3\text{M}}{4\pi\text{R}^3}=\frac{3\times2.0\times10^{30}}{4\times3.14(7\times10^8)^3}$

$=1.392\times10^3\text{Kg/m}^3$

The density of the Sun is in the density range of solids and liquids. This high density is attributed to the intense gravitational attraction of the inner layers on the outer layer of the Sun.

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Question 23 Marks

A physical quantity P is related to four observables a, b, c and d as follows:

$\text{P}=\text{a}^3\text{b}^3/\big(\sqrt{\text{c}}\text{ d}\big)$

The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2%, respectively. What is the percentage error in the quantity P ? If the value of P calculated using the above relation turns out to be 3.763, to what value should you round off the result ?

Answer

$\text{P}=\frac{\text{a}^3\text{b}^2}{(\sqrt{\text{c}}\text{ d})}$

Maximum fractional error in P is given by

$\frac{\Delta\text{P}}{\text{P}}=\pm\Big[3\frac{\Delta\text{a}}{\text{a}}+2\frac{\Delta\text{b}}{\text{b}}+\frac{1}{2}\frac{\Delta\text{c}}{\text{c}}+\frac{\Delta\text{d}}{\text{d}}\Big]\\=\pm\Big[3\Big(\frac{1}{100}\Big)+2\Big(\frac{3}{100}\Big)+\frac{1}{2}\Big(\frac{4}{100}\Big)+\frac{2}{100}\Big]$

$=\pm\frac{13}{100}=\pm0.13$

Percentage error in $\text{P}=\frac{\Delta\text{P}}{\text{P}}\times100=\pm0.13\times100=\pm13\%$

Percentage error in P = 13%

Value of P is given as 3.763.

By rounding off the given value to the first decimal place, we get P = 3.8.

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Question 33 Marks

A calorie is a unit of heat (energy in transit) and it equals about 4.2 J where 1J = 1kg m² s^–2. Suppose we employ a system of units in which the unit of mass equals $\alpha\text{ kg},$ the unit of length equals $\beta\text{ m},$ the unit of time is $\gamma\text{ s}.$ Show that a calorie has a magnitude $4.2\ \alpha^{-1}\ \beta^{-2}\ \gamma^2$ in terms of the new units.

Answer

Given that,

1 Calorie = 4.2 J = 4.2Kg m² s^-2 …… (i)

As new unit of mass $\alpha\text{ kg}$

$\therefore\ 1\text{kg}=1/\alpha$ new unit of mass

Similarly, $1\text{m}=\beta^{-1}$ new unit of length

$1\text{s}=\gamma^{-1}$ new unit of time

Putting these values in (i), we get

1 calorie = 4.2 ($\alpha^{-1}$ new unit of mass)($\beta^{-1}$ new unit of length)²($\gamma^{-1}$ new unit of time)^-2

$=4.2\ \alpha^{-1}\beta^{-2}\ \gamma^{2}$ new unit of energy (Proved)

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Question 43 Marks

Just as precise measurements are necessary in science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations. Think of ways by which you can estimate the following (where an estimate is difficult to obtain, try to get an upper bound on the quantity):
The number of air molecules in your classroom.

Answer

Let the volume of the room be V.

One mole of air at NTP occupies 22.4 l i.e., 22.4 × 10^-3 m³ volume.

Number of molecules in one mole = 6.023 × 10²³

$\therefore$ Number of molecules in room of volume V

= 6.023 × 10²³/22.4 × 10^-3V = 134.915 × 10²⁶ V

= 1.35 × 10²⁸V

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Question 53 Marks

A SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate objects under water. In a submarine equipped with a SONAR, the time delay between generation of a probe wave and the reception of its echo after reflection from an enemy submarine is found to be 77.0s. What is the distance of the enemy submarine? (Speed of sound in water = 1450m s^–1).

Answer

Let the distance between the ship and the enemy submarine be ‘S’.

Speed of sound in water = 1450m/s

Time lag between transmission and reception of Sonar waves = 77s

In this time lag, sound waves travel a distance which is twice the distance between the ship and the submarine (2S).

Time taken for the sound to reach the submarine = 1/2 × 77 = 38.5s

$\therefore$ Distance between the ship and the submarine (S) = 1450 × 38.5 = 55825m = 55.8km

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Question 63 Marks

When the planet Jupiter is at a distance of 824.7 million kilometers from the Earth, its angular diameter is measured to be 35.72” of arc. Calculate the diameter of Jupiter.

Answer

Distance of Jupiter from the Earth, D = 824.7 × 10⁶km = 824.7 × 10⁹m

Angular diameter = 35.72^“ = 35.72 × 4.874 × 10^-6rad

Diameter of Jupiter = d

Using the relation,

$\theta=\frac{\text{d}}{\text{D}}$

$\text{d}=\theta\text{ D}=824.7\times10^9\times35.72\times4.872\times10^{-6}$

$= 143520.76 \times 10^3\text{m} = 1.435 \times 10^5\text{Km}$

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Question 73 Marks

A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8min and 20s to cover this distance?

Answer

Distance between the Sun and the Earth = Speed of light × Time taken by light to cover the distance

Given that in the new unit, speed of light = 1 unit

Time taken, t = 8min 20s = 500s

$\therefore$ Distance between the Sun and the Earth = 1 × 500 = 500 units

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Question 83 Marks

The principle of ‘parallax’ in section 2.3.1 is used in the determination of distances of very distant stars. The baseline AB is the line joining the Earth’s two locations six months apart in its orbit around the Sun. That is, the baseline is about the diameter of the Earth’s orbit ≈ 3 × 10¹¹m. However, even the nearest stars are so distant that with such a long baseline, they show parallax only of the order of 1” (second) of arc or so. A parsec is a convenient unit of length on the astronomical scale. It is the distance of an object that will show a parallax of 1” (second of arc) from opposite ends of a baseline equal to the distance from the Earth to the Sun. How much is a parsec in terms of metres?

Answer

Diameter of Earth's orbit = 3 × 10¹¹m

Radius of Earth's orbit, r = 1.5 × 10¹¹m

Let the distance parallax angle be1” = 4.847 × 10^-6 rad.

Let the distance of the star be D.

Parsec is defined as the distance at which the average radius of the Earth's orbit subtends an angle of 1”.

$\therefore$ We have $\theta=\frac{\text{r}}{\text{D}}$

$\text{D}=\frac{\text{r}}{\theta}=\frac{1.5\times10^{11}}{4.847\times10^{-6}}$

$=0.309\times10^{-6}\approx3.09\times10^{16}\text{m}$

Hence, 1 parsec $\approx3.09\times10^{16}\text{m}.$

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Question 93 Marks

The photograph of a house occupies an area of 1.75cm² on a 35mm slide. The slide is projected on to a screen, and the area of the house on the screen is 1.55m². What is the linear magnification of the projector-screen arrangement.

Answer

Area of the house on the slide = 1.75cm²

Area of the image of the house formed on the screen = 1.55m²

= 1.55 × 10⁴cm²

Arial magnification, m_a = Area of Image/Area of Object = (1.55/1.75) × 10⁴

$\therefore$ Linear magnifications, m_l = underroot m_a

$=\sqrt{\frac{1.55}{1.75}\times10^4}=94.11$

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Question 103 Marks

Just as precise measurements are necessary in science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations. Think of ways by which you can estimate the following (where an estimate is difficult to obtain, try to get an upper bound on the quantity):
The total mass of rain-bearing clouds over India during the Monsoon.

Answer

During monsoons, a metrologist records about 215cm of rainfall in India i.e., the height of water column, h = 215cm = 2.15m
Area of country, A = 3.3 × 10¹²m²
Hence, volume of rain water, V = A × h = 7.09 × 10¹²m³
Density of water, p = 1 × 10³kg m^-3
Hence, mass of rain water = p × V = 7.09 × 10¹⁵kg
Hence, the total mass of rain-bearing clouds over India is approximately 7.09 × 10¹⁵kg

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Question 113 Marks

Explain this common observation clearly: If you look out of the window of a fast moving train, the nearby trees, houses etc. seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hill tops, the Moon, the stars etc.) seem to be stationary. (In fact, since you are aware that you are moving, these distant objects seem to move with you).

Answer

Line of sight is defined as an imaginary line joining an object and an observer’s eye. When we observe nearby stationary objects such as trees, houses, etc. while sitting in a moving train, they appear to move rapidly in the opposite direction because the line of sight changes very rapidly.
On the other hand, distant objects such as trees, stars, etc. appear stationary because of the large distance. As a result, the line of sight does not change its direction rapidly.

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Question 123 Marks

The length, breadth and thickness of a rectangular sheet of metal are 4.234m, 1.005m, and 2.01cm respectively. Give the area and volume of the sheet to correct significant figures.

Answer

Given that,
length, l = 4.234m
breadth, b = 1.005m
thickness, t = 2.01cm = 2.01 × 10^-2m
Area of the sheet = 2(l × 0 + b × t + t × l) = 2(4.234 × 1.005 + 1.005 × 0.0201 + 0.0201 × 4.234)
= 2(4.3604739) = 8.7209478 m2
As area can contain a maximum of three significant digits, therefore, rounding off, we get
Area = 8.72m²
Also, volume = l × b × t
V = 4.234 × 1.005 × 0.0201 = 0.0855289 = 0.0855m³(Significant Figures = 3)

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Question 133 Marks

Just as precise measurements are necessary in science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations. Think of ways by which you can estimate the following (where an estimate is difficult to obtain, try to get an upper bound on the quantity):
The mass of an elephant.

Answer

Consider a ship of known base area floating in the sea. Measure its depth in sea (say d₁).
Volume of water displaced by the ship, Vb = Ad₁
Now, move an elephant on the ship and measure the depth of the ship (d₂) in this case.
Volume of water displaced by the ship with the elephant on board, V_be = Ad₂
Volume of water displaced by the elephant = Ad₂ - Ad₁
Density of water = D
Mass of elephant = AD(d₂ - d₁)

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Question 143 Marks

Precise measurements of physical quantities are a need of science. For example, to ascertain the speed of an aircraft, one must have an accurate method to find its positions at closely separated instants of time. This was the actual motivation behind the discovery of radar in World War II. Think of different examples in modern science where precise measurements of length, time, mass etc. are needed. Also, wherever you can, give a quantitative idea of the precision needed.

Answer

It is indeed very true that precise measurements of physical quantities are essential for the development of science. For example, ultra-shot laser pulses (time interval ∼ 10^–15 s) are used to measure time intervals in several physical and chemical processes.
X-ray spectroscopy is used to determine the inter-atomic separation or inter-planer spacing.
The development of mass spectrometer makes it possible to measure the mass of atoms precisely.

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Question 153 Marks

The farthest objects in our Universe discovered by modern astronomers are so distant that light emitted by them takes billions of years to reach the Earth. These objects (known as quasars) have many puzzling features, which have not yet been satisfactorily explained. What is the distance in km of a quasar from which light takes 3.0 billion years to reach us?

Answer

Time taken by quasar light to reach Earth = 3 billion years
= 3 × 10⁹ years
= 3 × 10⁹ × 365 × 24 × 60 × 60s
Speed of light = 3 × 10⁸m/s
Distance between the Earth and quasar
= (3 × 10⁸) × (3 × 10⁹ × 365 × 24 × 60 × 60)
= 283824 × 10²⁰m
= 2.8 × 10²²km

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Question 163 Marks

Consider a simple pendulum, having a bob attached to a string, that oscillates under the action of the force of gravity. Suppose that the period
of oscillation of the simple pendulum depends on its length (l), mass of the bob (m) and acceleration due to gravity (g). Derive the expression for its time period using method of dimensions.

Answer

The dependence of time period $T$ on the quantities $l, g$ and $m$ as a product may be written as :
$
T=k P^r g^y m^z
$
where $k$ is dimensionless constant and $x, y$ and $z$ are the exponents.

By considering dimensions on both sides, we have
$
\begin{array}{l}
{\left[ L ^0 M ^0 T ^1\right]=\left[ L ^1\right]^x\left[ L ^1 T ^{-2}\right]^y\left[ M ^1\right]^z} \\
= L ^{x \neq y} T ^{2 y} M ^z
\end{array}
$

On equating the dimensions on both sides, we have
$
x+y=0 ;-2 y=1 \text {; and } z=0
$

So that $x=\frac{1}{2}, y=-\frac{1}{2}, z=0$
Then, $T=k\ l^\frac{1}{2} g^\frac{1}{2}$
or, $T=k \sqrt{\frac{l}{g}}$
Note that value of constant $k$ can not be obtained by the method of dimensions. Here it does not matter if some number multiplies the right side of this formula, because that does not affect its dimensions.
Actually, $k=2 \pi$ so that $T=2 \pi \sqrt{\frac{l}{g}}$

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Question 173 Marks

Why has second been defined in terms of periods of radiations from cesium-133?

Answer

Second has been defined in terms of periods of radiation, because

This period is accurately defined.
This period is not affected by change of physical conditions like temperature, pressure and volume etc.
The unit is easily reproducible in any good laboratory.

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Question 183 Marks

Find the dimensions of the following quantities

Acceleration.
Angle.
Density.
Kinetic energy.
Gravitational constant.
Permeability.

Answer

Acceleration $=\frac{\text{Velocity}}{\text{Time}}$

$\therefore[\text{Acceleration}]=\frac{[\text{Velocity}]}{[\text{Time}]}=\frac{[\text{LT}^{-1}]}{[\text{T}]}=[\text{LT}^{-2}]$

$\text{Angle}=\frac{\text{Distance}}{\text{Distance}}$

$\therefore$ angle is dimensionless

$\text{Density}=\frac{\text{Mass}}{\text{Volume}}$

$\therefore[\text{Density}]=\frac{[\text{Mass}]}{[\text{Volume}]}$

$\therefore[\text{Density}]=\frac{[\text{Mass}]}{[\text{Volume}]}=\frac{[\text{M}]}{[\text{L}^3]}=[\text{ML}^{-3}]$

Kinetic energy $=\frac{1}{2}\text{Mass}\times\text{Velocity}^2$

$\therefore$ [Kinetic energy] $=[\text{Mass}]\times[\text{Velocity}]^2=[\text{ML}^2\text{T}^{-2}]$

Constant of gravitation occurs in Newton's law of gravitation:

$\text{F}=\text{G}\frac{\text{m}_1\text{m}_2}{\text{d}^2}$

$\therefore[\text{G}]=\frac{[\text{F}][\text{d}^2]}{[\text{m}_1][\text{m}_2]}$

$=\frac{\text{MLT}^{-2}\text{L}^2}{[\text{M}][\text{M}]}=[\text{M}^{-1}\text{L}^3\text{T}^{-2}]$

Permeability occurs in Ampere's law of force:

$\Delta\text{F}=\mu\frac{(\text{i}_1\Delta\text{l}_1)(\text{i}_2\Delta\text{l}_2)\sin\theta}{\text{r}^2}$

$\therefore[\mu]=\frac{[\Delta\text{F}][\text{r}^2]}{[\text{i}_1\Delta\text{l}_1][\text{i}_2\Delta\text{l}_2]}$

$=\frac{[\text{MLT}^{-2}]}{[\text{AL}][\text{AL}]}=[\text{MLT}^{-2}\text{A}^{-2}]$

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Question 193 Marks

The wavelength 1 associated with a moving particle depends upon its mass m, its velocity v and Planck's constant h. Show dimensional relation between them.

Answer

Suppose wavelength à associated with a moving particle depends upon (i) its mass (m), (ii) its velocity (v) and (iii) Planck's constant (r)

where, k is a dimensionless constant.

Writing dimensions of various terms,

We get:

$[\text{M}^0\text{L}^1\text{T}^0]=[\text{M}]^{\text{a}}[\text{LT}^{-1}]^{\text{b}}[\text{ML}^2\text{T}^{-1}]^{\text{c}}$

$=\text{M}^{\text{a}+\text{c}}\text{L}^{\text{b}+\text{2c}}\text{T}^{-\text{b}-\text{c}}$

Comparing power of M, L and T on two sides of equation,

We have:

$\text{a}+\text{c}=0,\text{b}+\text{2c}=1,-\text{b}-\text{c}=0$

We get:

$\text{a}=-1,\text{b}=-1,\text{c}=+1$

Hence, the relation becomes $\lambda=\frac{\text{kh}}{\text{mv}}$

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Question 203 Marks

Compute the following with regards to significant figures.

$4.6\times0.128$
$\frac{0.9995\times1.53}{1.592}$
$876+0.4382$

Answer

$4.6 \times 0.128 = 0.5888 = 0.59$

The result has been rounded off to have two significant digits (as in 4.6)

$\frac{0.9995\times1.53}{1.592}=0.96057=0.961$
$876 + 0.4382 = 876.4382 = 876 .$

As, there is no decimal point in 876, therefore, result of addition has been rounded off to no decimal point.

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Question 213 Marks

Some measurements are taken for the period of oscillations of a simple pendulum. In successive measurements, the readings turn out to be 2.63s, 2.56s, 2.42s, 2.71s and 2.80s. Calculate the absolute errors, relative error and percentage error.

Answer

Mean value of time period,

$\text{T}=\frac{2.63+2.56+2.42+2.71+2.80}{5}$

$=2.62\text{sec}$

Absolute error in time period

$\Delta\overline{\text{T}}=\frac{0.01+0.06+0.20+0.09+0.18}{5}=0.11\text{ sec}$

Relative error, $\frac{\Delta\overline{\text{T}}}{\overline{\text{T}}}=\frac{0.11}{2.62}=0.04$

Percentage error, $\frac{\Delta\text{T}}{\text{T}}\times100=0.04\times100=4\%$

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Question 223 Marks

Check by the method of dimensional analysis whether the following relations are correct.
$\text{v}=\sqrt{\frac{\text{P}}{\text{D}}}$ where v = velocity of sound and P = pressure, D = density of medium
$\text{n}=\frac{1}{21}\sqrt{\frac{\text{F}}{\text{m}}},$ where n = frequency of vibration
l = legnth of the string
F = Stretching force
m = mass per unit length of the string.

Answer

$[\text{R.H.S}]=\sqrt{\frac{[\text{P}]}{\text{[D]}}}=\sqrt{\frac{[\text{ML}^{-1}\text{T}^{-2}]}{[\text{ML}^{-3}]}}=[\text{LT}^{-1}]$

$[\text{L.H.S}]=[\text{v}]=[\text{LT}^{-1}]$

$[\text{R.H.S.}]=[\text{L.H.S.}]$

Hence, the relation is correct.

$[\text{R.H.S.}]=\frac{1}{[\text{l}]}\sqrt{\frac{[\text{F}]}{[\text{m}]}}$

$=\frac{1}{\text{L}}\sqrt{\frac{\text{MLT}^{-2}}{\text{ML}^{-1}}}=\frac{1}{\text{L}}[\text{LT}^{-1}]=[\text{T}^{-1}]$

$[\text{L.H.S.}]=\Big[\frac{1}{\text{Time}}\Big]=\frac{1}{\text{T}}=[\text{T}^{-1}]$

Hence, the relation is correct.

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Question 233 Marks

The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 107K, and its outer surface at a temperature of about 6000K. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases? Check if your guess is correct from the following data : mass of the Sun = 2.0 × 1030kg, radius of the Sun = 7.0 × 108m.

Answer

Mass of the Sun, M = 2.0 × 10³⁰kg

Radius of the Sun, R = 7.0 × 10⁸m

Density, $\rho= ?$

$\rho=\frac{\text{mass}}{\text{volume}}=\frac{\text{M}}{\frac{4}{3}\pi\text{R}^3}=\frac{3\text{M}}{4\pi\text{R}^3}=\frac{3\times2.0\times10^{30}}{4\times3.14(7\times10^8)^3}$

$=1.392\times10^3\text{Kg/m}^3$

The density of the Sun is in the density range of solids and liquids. This high density is attributed to the intense gravitational attraction of the inner layers on the outer layer of the Sun.

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Question 243 Marks

Write the dimensional formula for the following:

Wein’s constant.
Planck's constant.
Specific heat.
Latent heat.
Rydberg's constant.

Answer

$[\text{M}^0\text{LT}^0\text{K}]$
$[\text{ML}^2\text{T}^{-1}]$
$[\text{M}^0\text{L}^{2}\text{T}^{-2}\text{K}^{-1}]$
$[\text{M}^0\text{L}^2\text{T}^{-2}]$
$[\text{M}^0\text{L}^{-1}\text{T}^0].$

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Question 253 Marks

In an experiment, refractive index of glass was observed to be 1.45, 1.56, 1.54, 1.44, 1.54 and 1.53.

Calculate:

Mean value of refractive index.
Mean absolute error.
Fractional error.
Percentage error.

Express the result in terms of absolute error and percentage error.

Answer

Mean value of $\mu$

$=\frac{1.45+1.56+1.54+1.44+1.54+1.53}{6}$

$\mu=1.51$

Mean absolute error $=\frac{\text{sum of absolute error}}{6}$

$\Delta\overline{\mu}=\frac{0.06+0.05+0.03+0.07+0.03+0.02}{6}$

$=\frac{0.26}{6}=0.0433=0.04$

Fractional error $=\frac{\Delta\overline{\mu}}{\mu}=\frac{0.04}{1.51}$

$=0.02649=0.03$

Percentage error $=\frac{\Delta\overline{\mu}}{\mu}\times100=3\%$

$\mu=1.51\pm0.04$ in terms of absolute error.

Also, $\mu=1.51\pm3\%$ in terms of % error.

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Question 263 Marks

Write the dimensions of:

Linear density.
Power.
Impulse.
Velocity gradien.
Mass per unit area.
Kinetic energy.
Angular acceleration.
Couple.
Moment of force.
Work done.

Answer

[ML^-1T⁰)
[ML²T^-3]
[MLT^-1)
[M⁰L⁰T^-1]
[ML^-2T⁰)
[M¹L²T^-2]
[M⁰L⁰T^-2]
[ML²T^-2]
[ML²T^-2]
[ML²T^-2].

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Question 273 Marks

The orbital velocity v of a satellite may depend on its mass m, distance r from the centre of Earth and acceleration due to gravity g. Obtain an expression for orbital velocity.

Answer

Let orbital velocity of satellite be given by the relation v = k m^ar^bg^c where k is a dimensionless constant and a, b and c are the unknown powers. Writing dimensions on two sides of equation, we have:

$[\text{M}^0\text{L}^1\text{T}^{-1}=[\text{M}]^{\text{a}}[\text{L}]^{\text{b}}[\text{LT}^{-2}]^{\text{c}}$

$=[\text{M}^{a}\text{L}^{\text{b}+\text{c}}\text{T}^{-2\text{c}}]$

Applying principle of homogeneity of dimensional equation,

We find that:

$\text{a}=0$

$\text{b}+\text{c}=1$

$-2\text{c}=-1$

On solving these equations,

We find that:

$\text{a}=0,\text{b}=+\frac{1}{2}\text{ and }\text{c}=+\frac{1}{2}$

$\therefore\text{v}=\text{k}\text{r}^{\frac{1}{2}}\text{g}^{\frac{1}{2}}$

Or $\text{v}=\text{k}\sqrt{\text{rg}}.$

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Question 283 Marks

The diameter of a wire as measured by a screw gauge was found to be 1.328, 1.330, 1.325, 1.334 and 1.336cm. Calculate

Mean value of diameter.
Absolute error in each measurement.
Mean absolute error.
Fractional error.
Percentage error.
Diameter of wire.

Answer

Mean value of diameter,

$\text{D}_\text{m}=\frac{1.328+1.330+1.325+1.326+1.334+1.336}{6}$

$=\frac{7.979}{6}=13298=1.330$

[rounding off to three decimal places]

Absolute error in different observations are:

$\Delta\text{D}_1=|1.330-1.328|=0.002\text{cm}$

$\Delta\text{D}_2=|1.330-1.330|=0\text{cm}$

$\Delta\text{D}_3=|1.330-1.325|=\pm0.005\text{cm}$

$\Delta\text{D}_4=|1.330-1.326|=0.004\text{cm}$

$\Delta\text{D}_5=|1.330-1.334|=\pm0.004\text{cm}$

$\Delta\text{D}_6=|1.330-1.336|=\pm0.006\text{cm}$

Mean absolute error

$\Delta\text{D}_{\text{mean}}=\frac{|\Delta\text{D}_1|+|\Delta\text{D}_2|+|\Delta\text{D}_3|+|\Delta\text{D}_\text{4}|+|\Delta\text{D}_5|+|\Delta\text{D}_6|}{6}$

$=\frac{0.002+0+0.005+0.004+0.004+0.004+0.006}{6}$

$=\frac{0.021}{6}=0.0035=0.004$ (rounding off to 3 decimal places)

Fractional error $=\frac{\Delta\text{D}_{\text{mean}}}{\text{D}}=\pm\frac{0.004}{1.330}=\pm0.003$
Percentage error $=\pm0.003\times100\%=\pm0.3\%$
Diameter of wire $=(1.330\pm0.03)\text{cm}$

Or $\text{D}=1.330\text{cm}\pm0.3\%$

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Question 293 Marks

Find the value of 60W on a system having 100g, 20cm and 1 minute as the fundamental units.

Answer

Here n₁ = 60W. Obviously, the physical quantity is power whose dimensional formula is

[M¹ L²T^-3]. The first system, in which unit of power is 1 watt, is SI system in which M = 1kg L₁ = 1m and T₁ = 1 s in second system, M₂ = 100g, L₂ = 20cm and T₂ = 1 min = 60s.

$\therefore\text{n}_2=\text{n}_1\Big[\frac{\text{M}_1}{\text{M}_2}\Big]^1\Big[\frac{\text{L}_1}{\text{L}_2}\Big]^2\Big[\frac{\text{T}_1}{\text{T}_2}\Big]^{-3}$

$=60\Big[\frac{1\text{kg}}{100\text{g}}\Big]^1\Big[\frac{1\text{m}}{20\text{cm}}\Big]^2\Big[\frac{1\text{s}}{1\text{min}}\Big]^{-3}$

$=60\Big[\frac{1000\text{g}}{100\text{g}}\Big]^1\Big[\frac{100\text{cm}}{20\text{cm}}\Big]^2\Big[\frac{1\text{s}}{60\text{s}}\Big]^{-3}$

$=60\times\frac{1000}{100}\times\frac{100}{20}\times\frac{100}{20}\times60\times60\times60$

$=3.24\times10^9\text{units}$

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Question 303 Marks

Write the dimensions of the following:

Gravitational potential.
Variable force.
Pressure gradient.
Moment of inertia.
Buoyant force.
Angular momentum.
Work done by torque.
Moment of momentum.
Moment of force.
Pressure energy.

Answer

[M⁰L²T^-2]
[MLT^-2]
[ML^-2T^-2]
[ML²T⁰]
[MLT^-2]
[ML²T^-1]
[ML²T^-2)
[ML²T^-2]
[ML²T^-2]
[ML²T^-2]

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Question 313 Marks

Einstein’s mass - energy relation emerging out of his famous theory of relativity relates mass (m) to energy (E) as E = mc², where c is speed of light in vacuum. At the nuclear level, the magnitudes of energy are very small. The energy at nuclear level is usually measured in MeV, where 1MeV = 1.6 × 10^-13J; the masses are measured in unified atomic mass unit (u) where 1u = 1.67 × 10^-27kg.

Show that the energy equivalent of 1u is 931.5MeV.
A student writes the relation as 1u = 931.5MeV. The teacher points out that the relation is dimensionally incorrect. Write the correct relation.

Answer

We can apply Einstein’s mass-energy relation in this problem, E = mc², to calculate the energy equivalent of the given mass.

Here, $1\ \text{amu}=1\text{u}=1.67\times10^{-27}\text{kg}$

Applying E = mc²

Energy $\text{E}=(1.67\times10^{-27})(3\times10^8)\text{J}=1.67\times9\times10^{-11}\text{J}$

$\text{E}=\frac{1.67\times9\times10^{-11}}{1.6\times10^{-13}}\text{MeV}$

$=939.4\text{MeV}\approx931.5\text{MeV}$

$\text{As E}=\text{mc}^2\Rightarrow\text{m}=\frac{\text{E}}{\text{c}^2}$

According to this, $1\text{u}=\frac{931.5\text{MeV}}{\text{c}^2}$

Hence the dimensionally correct relation $1\ \text{amu}\times\text{c}^2=1\text{u}\times\text{c}^2=931.5\text{MeV}$

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Question 323 Marks

The distance of the Sun from the Earth is 1.496 × 10¹¹ m (i.e., 1 A.U.). If the angular diameter of the Sun is 2000”, find the diameter of the Sun.

Answer

Here, $\theta=2000''=\frac{2000}{3600}\times\frac{\pi}{180}\text{rad}$

$=9.7\times10^{-3}\text{rad}$

$\text{d}=1.496\times10^{11}\text{m}$

From the figure,

$\theta=\frac{\text{D}}{\text{d}}$

$\therefore\text{D}=\theta\text{d}$

$=9.7\times10^{-3}\times1.496\times{10}^{11}$

$=1.45\times10^{9}\text{m}$

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Question 333 Marks

By using the method of dimension, check the accuracy of the following formula : $ \text{T}=\frac{\text{rh}\rho\text{g}}{2\cos\theta},$where T is the surface tension, h is the height of the liquid,$\rho$ is the density of the liquid, g acceleration due to gravity $\theta$ angle of contact, and r is the radius of the tube.

Answer

In order to find out the accuracy of the given equation we shall compare the dimensions of T and $\frac{\text{rh}\rho\text{g}}{2\cos\theta}$

$\text{T}=\frac{\text{force}}{\text{length}}=\frac{[\text{MLT}^{-2}]}{[\text{L}]}$

Dimension of $\frac{\text{rh}\rho\text{g}}{2\cos\theta}=[\text{L][L][ML}^{-3}][\text{LT}^{-2}]$

$(2\cos\theta$ has no dimension$)$

The dimensions of both the sides are the same and hence the equation is correct.

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Question 343 Marks

A physical quantity Q is given by

$\text{Q}= \frac{\text{A}^2\text{B}^\frac{3}{2}}{\text{C}^{+4}\text{D}^\frac{1}{2}}$

The percentage error in A, B, C, D are 1%, 2%, 4%, 2% respectively. Find the percentage error in Q.

Answer

$\% \text{ error in} \text{ Q}=2 \Big(\frac{\text{dA}}{A}\times100\Big)+\frac{3}{2}\Big(\frac{\text{dB}}{\text{B}}\times100\Big)$
$+4\Big(\frac{\text{dC}}{\text{C}}\times100\Big)+\frac{1}{2}\Big(\frac{\text{dB}}{\text{D}}\times100\Big) $
$= 2\times1+\frac{3}{2}\times2+4\times4+\frac{1}{2}\times2$
$=2+3+16+1= 22\%$

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Question 353 Marks

Express unified atomic mass unit in kg.

Answer

The unified atomic mass unit is the standard unit that is used for indicating mass on an atomic or molecular scale (atomic mass). One unified atomic mass unit is approximately the mass of one nucleon (either a single proton or neutron) and is numerically equivalent to 1g/mol. It is defined as one- twelfth of the mass of an unbound neutral atom of carbon-12 in its nuclear and electronic ground state.

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Question 363 Marks

Experiments show that the frequency (n) of a tuning fork depends upon the length (I) of the prong, the density (d) and the Young's modulus (Y) of its material. From dimensional considerations, find a possible relation for the frequency of the tuning fork.

Answer

$\text{Let n }= \text{Kl}^\text{a}\text{d}^\text{b}\text{y}^\text{c}, $
Where $\text{K}$
Substituting the dimension of all the quantities involved
We have: $[\text{T}^{-1}]= [\text{L}]^\text{a}[\text{ML}^{-3}]^\text{b}[\text{ML}^{-1}\text{T}^{-2}]^\text{c}$
$ [\text{M}^0\text{L}^0\text{T}^{-1}]= [\text{M}] ^{(\text{b+c})}[\text{L}]^{\text{a-3b-c}}[\text{T}] ^{-2\text{c}}$
Comparing powers of M, L and T we get
$\text{b+c = 0}$
$\text{a}-3\text{b}-\text{c}=0$
$-2\text{c}= -1$
Or $\text{a = }-1, \text{b}= \frac{-1}{2} \text{ and c} =\frac{1}{2}$
This gives $\text{n}= \text{Kl}^{-1}\text{d}^\frac{-1}{2}\text{y}\frac{1}{2}$
Or $\text{n}= \frac{\text{K}}{1}\sqrt{\frac{\text{y}}{\text{d}}}$

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Question 373 Marks

In the expression P = El² m^-5 G^-2, E, m, l and G denote energy, mass, angular momentum and gravitational constant, respectively. Show that P is a dimensionless quantity.

Answer

According to the problem, expression is $\text{P}=\text{El}^2\text{m}^{-5}\text{G}^{-2}$
where E is energy $[\text{E}]=[\text{ML}^2\text{T}^{-2}],$ m is mass [m] = [M],
L is angular momentum $[\text{L}] = [\text{ML}^2 \text{T}^{-1}],$ G is gravitational constant $[\text{G}] = [\text{M}^{-1}\text{L}^2\text{T}^{-2}]$
Substituting dimensions of each physical quantity in the given expression,
$[\text{P}]=[\text{ML}^2\text{T}^{-2}]\times[\text{ML}^2\text{T}^{-1}]^2\times[\text{M}]^{-5}\times[\text{M}^{-1}\text{L}^3\text{T}^{-2}]^{-2}$
$=[\text{M}^{1+2-5+2}\text{L}^{2+4-6}\text{T}^{-2-2+4}]$
$=[\text{M}^0\text{L}^0\text{T}^0]$
This shows that P is a dimensionless quantity.

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Question 383 Marks

Reynold's number N_R(a dimensionless quantity) determines the condition of laminar flow of a viscous liquid through a pipe. N_R is a function of the density of the liquid 'r', its average speed is 'y' and the coefficient of viscosity of the liquid is 'h'. If N, is given directly proportional to 'd' (the diameter of the pipe), show from dimensional consideration that $\text{N}_\text{R}\propto \frac{\text{dp}\rho}{\eta}$ the unit of '$\eta$' in SI system is kg m^-1s^-1?

Answer

As the Reynold's Number N_R depends on density p, average speed v and coefficient ofviscosity I, then let us say.

$\text{N}_\text{R}\propto \rho^\text{a}\text{v}^b\eta^\text{c}$

Again is proportional to the diameter of the pipe , combining the two we have . $\text{N}_\text{R}\propto \rho ^\text{a}\text{v}^\text{b}\eta^\text{c}\text{d}$

$\text{N}_\text{R }= \text{ K} \rho^\text{a} \text{v}^\text{b}\eta^\text{c}\text{d}$

We write ,$[\text{N}_\text{R}]= [\text{M}^0\text{L}^0\text{T}^0]$

$[\rho]= [\text{MK}^{-3}]$

$[\text{v}]= [\text{LT}^{-1}]$

$[\eta] = [ \text{ML}^{-1} \text{T}^{-1}]$

$[\text{d}]= [\text{L}]$

Syubstituting the dimension in (i), we have ,

$[\text{M}^0\text{L}^0\text{T}] = [\text{ML}^{-3}]^\text{a}[\text{LT}^{-1}]^\text{b}[\text{ML}^{-1}\text{T}^{-1}]^\text{c}[\text{L}]$

$= \text{M}^{(\text{a+c})} \text{L }^{(-3\text{a+b+c+1})}\text{T}^{(\text{-b-c})}$

Comparing the dimensions of M,L and T, we have,

$\text{a+c}=0$

$-3\text{a+b-c+1=0}$

$-\text{b}-\text{c} =0$

On simplifying, we get

$\text{c}= -1$

$\text{b}=1$

$\text{a}=1$

Therefore, the relation(i) becomes

$\text{N}_\text{R} =\text{K}\rho^1 \text{v}^1\eta^{-1}\text{d}$

$\text{N}_\text{R}=\text{K}\rho^1\text{v}^1\eta^{-1}\text{d}$

$\text{N}_\text{R} = \text{K}\rho\frac{\text{vd}}{\eta}$

$\text{N}_\text{R} = \propto\rho\frac{\text{vd}}{\eta}$

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Question 393 Marks

Why length, mass and time are chosen as base quantities in mechanics?

Answer

Normally each physical quantity requires a unit or standard for its specification, so it appears that there must be as many units as there are physical quantities. However, it is not so. It has been found that if in mechanics we choose arbitrarily units of any three physical quantities we can express the units of all other physical quantities in mechanics in terms of these. So, length, mass and time are chosen as base quantities in mechanics because.

Length, mass and time cannot be derived from one another, that is these quantities are independent.
All other quantities in mechanics can be expressed in terms of length, mass and time.

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Question 403 Marks

Just as precise measurements are necessary in science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations. Think of ways by which you can estimate the following (where an estimate is difficult to obtain, try to get an upper bound on the quantity):
The total mass of rain-bearing clouds over India during the Monsoon.

Answer

During monsoons, a metrologist records about 215cm of rainfall in India i.e., the height of water column, h = 215cm = 2.15m
Area of country, A = 3.3 × 10¹²m²
Hence, volume of rain water, V = A × h = 7.09 × 10¹²m³
Density of water, p = 1 × 10³kg m^-3
Hence, mass of rain water = p × V = 7.09 × 10¹⁵kg
Hence, the total mass of rain-bearing clouds over India is approximately 7.09 × 10¹⁵kg

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Question 413 Marks

Explain this common observation clearly: If you look out of the window of a fast moving train, the nearby trees, houses etc. seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hill tops, the Moon, the stars etc.) seem to be stationary. (In fact, since you are aware that you are moving, these distant objects seem to move with you).

Answer

Line of sight is defined as an imaginary line joining an object and an observer’s eye. When we observe nearby stationary objects such as trees, houses, etc. while sitting in a moving train, they appear to move rapidly in the opposite direction because the line of sight changes very rapidly.
On the other hand, distant objects such as trees, stars, etc. appear stationary because of the large distance. As a result, the line of sight does not change its direction rapidly.

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Question 423 Marks

The density of a cylindrical rod was measured by using the formula:
$\rho=\frac{4\text{m}}{\pi\text{D}^2\text{l}}$

Answer

The percentage errors in m, D and l are 1%, 1.5% and 0.5%. Calculate the percentage error in the calculated value of density.
$\text{Density }\rho=\frac{4\text{m}}{\pi\text{D}^2\text{l}}$
$\therefore\Big(\frac{\Delta\rho}{\rho}\Big)_{\text{max}}=\frac{\Delta\text{m}}{\text{m}}+2\frac{\Delta\text{D}}{\text{D}}+\frac{\Delta\text{l}}{\text{l}}$
But, $\frac{\Delta\text{m}}{\text{m}}=1\%,\frac{\Delta\text{D}}{\text{D}}=1.5\%\text{ and }\frac{\Delta\text{l}}{\text{l}}=0.5\%$
$\therefore$ Maximum percentage error in calculated value of density
$\Big(\frac{\Delta\rho}{\rho}\Big)_{\text{max}}=1\%+2\times1.5\%+0.5\%$
$=(1+3+0.5)\%=4.5\%$

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Question 433 Marks

The radius of curvature of a concave mirror measured by spherometer is given by $\text{R}=\frac{\text{I}^2}{6\text{h}}+\frac{\text{h}}{2}.$ The values of land h are 4cm and 0.065cm respectively. Compute the error in measurement of radius of curvature.

Answer

Here l = 4cm, $\Delta\text{I}=0.1\text{cm}$
(least count of the metre scale)
Here l is the distance between the legs of the spherometer. h = 0.065cm, Ah = 0.001cm = (least count of the spherometer)
$\text{As }\text{R}=\frac{\text{I}^2}{6\text{h}}+\frac{\text{h}}{2}$
Considering the magnitude only,
We get:
$\frac{\Delta\text{R}}{\text{R}}=2\frac{\Delta\text{I}}{\text{I}}+\frac{\Delta\text{h}}{\text{h}}+\frac{\Delta\text{h}}{\text{h}}$
$=2(\frac{\Delta\text{I}}{\text{I}}+\frac{\Delta\text{h}}{\text{h}})$
$=2\times\frac{0.1}{4}+\frac{2\times0.001}{0.065}$
$=0.05+0.03=0.08\%$

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Question 443 Marks

The sides of a rectangle are $(10.5\pm0.2)\text{cm}$ and $(52\pm0.1)\text{m}$ Calculate its perimeter with error limits.

Answer

$\text{Given},\text{l}=(10.5\pm0.2)\text{cm},$
$\text{b}=(5.2\pm0.1)\text{cm}$
Perimeter of a rectangle $\text{p}=2(\text{l}+\text{b})$
$=2(10.5+5.2)=3.14\text{cm}$
$\Delta\text{p}=\pm(\Delta\text{l}+\Delta\text{b})$
$=\pm2(0.2+0.1)=\pm0.6$
Perimeter of a rectangle $=(31.4\pm0.6)\text{cm}$

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Question 453 Marks

If the velocity of light (c), the constant of gravitation (G) and Planck's constant (h) be chosen as the fundamental units, find the dimensions of mass, length and time in the new system.

Answer

Let us write the dimensions of c, G and h in terms of M, L and T.
$\text{Let}\text{M}=\text{Kc}^a\text{G}^b\text{h}^\text{y},$ where K is constant
$\text{[M]}=\text{[LT}^{-1}]^\alpha\text{[M}^{-1}\text{[L}^3\text{T}^{-2}]^\beta\text{[ML}^2\text{T}^{-1}]^\gamma$
$=\text{[M}^{-\beta+\gamma}\text{L}^{\alpha+3\beta+2\gamma}\text{T}^{-\alpha-2\gamma}]$
Comparing the powers of M, L and T on both the sides, we have:
$-\beta+\gamma=1$
$\alpha+3\beta+2\gamma=0$
$-\alpha-2\beta-\gamma=0$
Solving these equations, we get,
$\alpha=\frac{1}{2},\beta=\frac{-1}{2},\gamma=\frac{1}{2}$
$\therefore \text{M}=\text{Kc}^\frac{1}{2}\text{G}^\frac{1}{2}\text{h}^\frac{1}{2}$
Taking$\text{K}=\text{I},$ we can write
$\text{M}=_\text{c}^\frac{1}{2}\text{G}^{\frac{-1}{2}}\text{h}^\frac{1}{2}$
Similary,we can prove that.
$\text{L}=_\text{c}^\frac{-3}{2}\text{G}^\frac{1}{2}\text{h}^\frac{1}{2}$
$\text{T}=_\text{c}^\frac{-5}{2}\text{h}^\frac{1}{2}\text{G}^\frac{1}{2}$

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Question 463 Marks

For a glass prism of refracting angle 60°, the minimum angle of deviation D_m is found to be 36° with a maximum error of 1.05°. When a beam of parallel light is incident on the prism, find the range of experimental value of refractive index $'μ'.$ It is known that the refractive index $'μ'$ of the material of the prism is given by:
$\mu=\frac{\sin\Big(\frac{\text{A}+\text{D}_\text{m}}{2}\Big)}{\sin\frac{\text{A}}{2}}$

Answer

Error in calculated D_m is $\pm1.05^{\circ}$
Given $\mu=\frac{\sin\Big(\frac{\text{A}+\text{D}_\text{m}}{2}\Big)}{\sin\frac{\text{A}}{2}}$
$\mu=\frac{\sin\Big(\frac{36^{\circ}\pm1.05}{2}}{\sin\Big(\frac{60}{2}\Big)}=\frac{\sin\Big(\frac{37.05}{2}\Big)}{\sin30^{\circ}}$
$=\frac{\sin(18.525)}{\frac{1}{2}}\text{or}\frac{\sin(17.475^{\circ})}{\frac{1}{2}}$
$\Rightarrow2\times0.755\text{ or }2\times0.73$
$\Rightarrow1.51\text{ or }1.46$
Here range of $\mu$ is $1.46\leq\mu\leq1.51$

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Question 473 Marks

The volume of a liquid flowing out per second of a pipe of length l and radius r is written by a student as,
$\text{v}=\frac{\pi}{8}\frac{\text{Pr}^4}{\eta\text{l}}$
where P is the pressure difference between the two ends of the pipe and $\eta$ is coefficent of viscosity of the liquid having dimensional formula ML^-1 T^-1.
Check whether the equation is dimensionally correct.

Answer

If dimensions of LHS of an equation is equal to dimensions of RHS, then equation is said to be dimensionally correct.
According to the problem, the volume of a liquid flowing out per second of a pipe is given by $\text{V}=\frac{\pi}{8}\frac{\text{pr}^2}{\eta\text{l}}$
(where, V = rate of volume of liquid per unit time)
Dimension of given physical quantities,
$[\text{V}]=\frac{\text{Dimension of volume}}{\text{Dimension of time}}=\frac{[\text{L}^3]}{[\text{T}]}=[\text{L}^3\text{T}^{-1}],[\text{p}]=[\text{ML}^{-1}\text{T}^{-2}],$
$[\eta]=[\text{ML}^{-1}\text{T}^{-1}],[\text{l}]=[\text{L}],[\text{r}]=[\text{L}]$
$\text{LHS}=[\text{V}]=\frac{[\text{L}^3]}{[\text{T}]}=[\text{L}^3\text{T}^{-1}]$
$\text{RHS}=\frac{[\text{ML}^{-1}\text{T}^{-2}]\times[\text{L}^4]}{[\text{ML}^{-1}\text{T}^{-1}]\times[\text{L}]}=[\text{L}^3\text{T}^{-1}]$
Dimensionally, L.H.S. = R.H.S.
Therefore, equation is correct dimensionally.

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Question 483 Marks

Express the average distance of earth from the sun in:

Light year.
Par sec.

Answer

Average distance of earth from the sun is (r)
$\frac{1.496\times10^{11}}{9.46\times10^{15}}=1.58\times10^{-5}\text{ ly}$
Also, $\text{r}=\frac{1.496\times10^{11}}{3.08\times10^{16}}\text{per sec}$
$=4.86\times10^{-6}\text{per sec}$

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Question 493 Marks

The radius of curvature of a concave mirror measured by spherometer is $\text{R}=\frac{\text{l}^2}{6\text{h}}+\frac{\text{h}}{2}$ The values of 1 and h are 4cm and 0.065cm respectively. Compute the error in measurement of radius of curvature.

Answer

We are given:
$\text{l}=4\text{cm},\Delta\text{l}=0.1\text{cm}$
(least count of the metre scale)
here l is the distance between the legs of the spherometer.
$\text{R}=\frac{\text{l}^2}{6\text{h}}+\frac{\text{h}}{2}$
$\therefore\frac{\Delta\text{R}}{\text{R}}=\frac{2\Delta\text{l}}{\text{l}}+\Big(-\frac{\Delta\text{h}}{\text{h}}\Big)+\frac{\Delta\text{h}}{\text{h}}$
Considering the magnitudes only,
We get:
$\frac{\Delta\text{R}}{\text{R}}=2\frac{\Delta\text{l}}{\text{l}}+\frac{\Delta\text{h}}{\text{h}}+\frac{\Delta\text{h}}{\text{h}}$
$=2\Big(\frac{\Delta\text{l}}{\text{l}}+\frac{\Delta\text{h}}{\text{h}}\Big)$
$=2\times\frac{0.1}{4}+\frac{2\times0.001}{0.065}=0.05+0.03=0.08.$

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Question 503 Marks

Just as precise measurements are necessary in science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations. Think of ways by which you can estimate the following (where an estimate is difficult to obtain, try to get an upper bound on the quantity):
The number of air molecules in your classroom.

Answer

Let the volume of the room be V.
One mole of air at NTP occupies 22.4 l i.e., 22.4 × 10^-3 m³ volume.
Number of molecules in one mole = 6.023 × 10²³
$\therefore$ Number of molecules in room of volume V
= 6.023 × 10²³/22.4 × 10^-3V = 134.915 × 10²⁶ V
= 1.35 × 10²⁸V

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3 Marks Question - Physics STD 11 Science Questions - Vidyadip